Answer:
Option C
Explanation:
Concept involved intersection of circles , the basic concept is to equations simulataneously and using properties of modules of complex numbers.
Formula used |z|2=z.\overline{z}
and |z_{1}-z_{2}|^{2}=(z_{1}-z_{2})(\overline{z_{1}}-\overline{z_{2}})
= |z_{1}|^{2}-z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}+|z_{2}|^{2}
Here,
(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2} and
(x-x_{0})^{2}+(y-y_{0})^{2}=4r^{2} could be written as,
|z-z_{0}|^{2}=r^{2} and |z-z_{0}|^{2}=4r^{2}
since ,\alpha and \frac{1}{\overline{\alpha}} lies on first and second respectively,
\therefore |\alpha-z_{0}|^{2}=r^{2} and |\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}
\Rightarrow (\alpha-z_{0})(\overline{\alpha}-\overline{z_{0}})=r^{2}
\Rightarrow |\alpha|^{2}-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|z_{0}|^{2}=r^{2} .....(i)
and |\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}
\Rightarrow (\frac{1}{\overline{\alpha}}-z_{0})(\frac{1}{\alpha}-\overline{z_{0}})=4r^{2}
\Rightarrow \frac{1}{|\alpha|^{2}}-\frac{z_{0}}{\alpha}-\frac{\overline{z_{0}}}{\overline{\alpha}}+|z_{0}|^{2}=4r^{2}
Since |\alpha|^{2}=\alpha.\alpha
\Rightarrow \frac{1}{|\alpha|^{2}}-\frac{z_{0}.\overline{\alpha}}{|\alpha|^{2}}-\frac{\overline{z_{0}}}{|{\alpha}|^{2}}\alpha+|z_{0}|^{2}=4r^{2}
\Rightarrow 1-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|\alpha|^{2}|z_{0}|^{2}=4r^{2}|\alpha|^{2} ...(ii)
On substracting Eqs (i) and (ii) , we get
(|\alpha|^{2}-1)+|z_{0}|^{2}(1-|\alpha|^{2})
= r^{2}(1-4|\alpha|^{2})
\Rightarrow (|\alpha|^{2}-1)(1-|z_{0}|^{2})=r^{2}(1-4|\alpha|^{2})
\Rightarrow (|\alpha|^{2}-1)\left(1-\frac{r^{2}+2}{2}\right)
=r^{2}(1-4|\alpha|^{2})
Given, |z_{0}|^{2}=\frac{r^{2}+2}{2}
\Rightarrow (|\alpha|^{2}-1).\left(\frac{-r^{2}}{2}\right)=r^{2}(1-4|\alpha|^{2})
\Rightarrow |\alpha|^{2}-1=-2+8|\alpha|^{2}
\Rightarrow 7|\alpha|^{2}=1
\therefore |\alpha|=\frac{1}{\sqrt{7}}